# What it the expected number of girls in a family of 3 children?

I think that I am doing it right, but I am not sure. Out of 3 children there are 7 possibilities of having a girl out of 8 possible outcomes. I am dividing the 7 possibilities of having a girl by 3 children: 7/3=2.33. I am not real good at this, therefore, is the expected number of girls in a family of 3 the 2.3 I came up with or am I doing this completely wrong? Thank you for any help that I can get!

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It cannot be right. If it were, the expected number of boys would also be $7/3$, giving an expected family size of $14/3$! –  André Nicolas Dec 3 '12 at 0:45

The family could have $0,1,2$, or $3$ girls. Of the $8$ possible sequences, only one gives them $0$ girls, so the probability of $0$ girls is $\frac18$. There are $3$ ways for them to have $1$ girl, however: GBB, BGB, and BBG. Thus, the probability of $1$ girl is $\frac38$. You can check similarly that they have $2$ girls with probability $\frac38$ and $3$ girls with probability $\frac18$.

The expected number of girls is a weighted average of the possible numbers, where the weight of a given number is its probability of occurring. Thus, the expected number of girls is

$$\frac18\cdot0+\frac38\cdot1+\frac38\cdot2+\frac18\cdot3=\frac{12}8=1.5\;.$$

You can also think of this as the average over all of the $8$ equally likely birth orders of three children: add up the number of girls in BBB, BBG, BGB, GBB, BGG, GBG, GGB, and GGG, and divide by the $8$ possibilities to get (again) $\frac{12}8=1.5$.

Finally, you can observe that since boys and girls are equally likely, the expected number of boys should be the same as the expected number of girls. Since each family has $3$ children altogether, and we expect half of them on average to be girls, we expect on average $\frac32=1.5$ girls per family.

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Each child has a $\frac12$ probability of being a girl. This is the expected number of girls among each one of the three children. Expectations are additive, so the expected number of girls among all three children is $\frac12+\frac12+\frac12 = \frac32 = 1.5$.

Similarly, the expected number of spots on one die is 3.5, so the expected number of spots on two dice added together is 3.5 + 3.5 = 7.

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